the sum of two consecutive even integers is at most seven more than half the sum of the next two consecutive even integers
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x<=10
To solve this problem, let's break it down step by step.
Let's assume the first even integer is x. The next consecutive even integer would then be x + 2.
Therefore, we can write the sum of the two consecutive even integers as x + (x + 2).
Similarly, let's assume the next two consecutive even integers are y and y + 2.
So, the sum of those two consecutive even integers can be written as y + (y + 2).
According to the problem statement, the sum of the first set of consecutive even integers is at most seven more than half the sum of the second set of consecutive even integers. Mathematically, we can represent this as:
x + (x + 2) ≤ (1/2)*(y + (y + 2)) + 7
Now, let's simplify this equation and solve for x:
2x + 2 ≤ (1/2)*(2y + 2) + 7
2x + 2 ≤ y + 1 + 7
2x + 2 ≤ y + 8
2x ≤ y + 6
x ≤ (y + 6)/2
So, the first even integer (x) is less than or equal to half of the sum of the second set of consecutive even integers plus 6 ([(y + 6)/2]).
This equation represents the relationship between x and y based on the given conditions.
Please note that we have not found any specific values for x or y, but we have expressed the relationship between them as requested.
x + x+2 <= 7+(x+4+x+6)/2
2x+2<= 7+x +5
Can you finish from here?